Problem: The lifespans of bears in a particular zoo are normally distributed. The average bear lives $36.4$ years; the standard deviation is $9$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a bear living longer than $18.4$ years.
Solution: $36.4$ $27.4$ $45.4$ $18.4$ $54.4$ $9.4$ $63.4$ $95\%$ $2.5\%$ $2.5\%$ We know the lifespans are normally distributed with an average lifespan of $36.4$ years. We know the standard deviation is $9$ years, so one standard deviation below the mean is $27.4$ years and one standard deviation above the mean is $45.4$ years. Two standard deviations below the mean is $18.4$ years and two standard deviations above the mean is $54.4$ years. Three standard deviations below the mean is $9.4$ years and three standard deviations above the mean is $63.4$ years. We are interested in the probability of a bear living longer than $18.4$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $95\%$ of the bears will have lifespans within 2 standard deviations of the average lifespan. The remaining $5\%$ of the bears will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({2.5\%})$ will live less than $18.4$ years and the other half $({2.5\%})$ will live longer than $54.4$ years. The probability of a particular bear living longer than $18.4$ years is ${95\%} + {2.5\%}$, or $97.5\%$.